3.951 \(\int \frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=83 \[ \frac{x \sqrt{c x}}{\sqrt [4]{a+b x^2}}+\frac{\sqrt{a} \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{b} \sqrt [4]{a+b x^2}} \]

[Out]

(x*Sqrt[c*x])/(a + b*x^2)^(1/4) + (Sqrt[a]*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a
]]/2, 2])/(Sqrt[b]*(a + b*x^2)^(1/4))

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Rubi [A]  time = 0.0324452, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {314, 284, 335, 196} \[ \frac{x \sqrt{c x}}{\sqrt [4]{a+b x^2}}+\frac{\sqrt{a} \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{b} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]/(a + b*x^2)^(1/4),x]

[Out]

(x*Sqrt[c*x])/(a + b*x^2)^(1/4) + (Sqrt[a]*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a
]]/2, 2])/(Sqrt[b]*(a + b*x^2)^(1/4))

Rule 314

Int[Sqrt[(c_)*(x_)]/((a_) + (b_.)*(x_)^2)^(1/4), x_Symbol] :> Simp[(x*Sqrt[c*x])/(a + b*x^2)^(1/4), x] - Dist[
a/2, Int[Sqrt[c*x]/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 284

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[(Sqrt[c*x]*(1 + a/(b*x^2))^(1/4))/(b*(a +
b*x^2)^(1/4)), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{\sqrt{c x}}{\sqrt [4]{a+b x^2}} \, dx &=\frac{x \sqrt{c x}}{\sqrt [4]{a+b x^2}}-\frac{1}{2} a \int \frac{\sqrt{c x}}{\left (a+b x^2\right )^{5/4}} \, dx\\ &=\frac{x \sqrt{c x}}{\sqrt [4]{a+b x^2}}-\frac{\left (a \sqrt [4]{1+\frac{a}{b x^2}} \sqrt{c x}\right ) \int \frac{1}{\left (1+\frac{a}{b x^2}\right )^{5/4} x^2} \, dx}{2 b \sqrt [4]{a+b x^2}}\\ &=\frac{x \sqrt{c x}}{\sqrt [4]{a+b x^2}}+\frac{\left (a \sqrt [4]{1+\frac{a}{b x^2}} \sqrt{c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{a x^2}{b}\right )^{5/4}} \, dx,x,\frac{1}{x}\right )}{2 b \sqrt [4]{a+b x^2}}\\ &=\frac{x \sqrt{c x}}{\sqrt [4]{a+b x^2}}+\frac{\sqrt{a} \sqrt [4]{1+\frac{a}{b x^2}} \sqrt{c x} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{b} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0098727, size = 56, normalized size = 0.67 \[ \frac{2 x \sqrt{c x} \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )}{3 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*x]/(a + b*x^2)^(1/4),x]

[Out]

(2*x*Sqrt[c*x]*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 3/4, 7/4, -((b*x^2)/a)])/(3*(a + b*x^2)^(1/4))

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{cx}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)/(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(1/2)/(b*x^2+a)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x)/(b*x^2 + a)^(1/4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral(sqrt(c*x)/(b*x^2 + a)^(1/4), x)

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Sympy [C]  time = 1.05564, size = 44, normalized size = 0.53 \begin{align*} \frac{\sqrt{c} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(1/2)/(b*x**2+a)**(1/4),x)

[Out]

sqrt(c)*x**(3/2)*gamma(3/4)*hyper((1/4, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(1/4)*gamma(7/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x}}{{\left (b x^{2} + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate(sqrt(c*x)/(b*x^2 + a)^(1/4), x)